The corners of regular tetrahedrons are numbe | vedclass

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(C) Each tetrahedron has $4$ faces,so when $3$ tetrahedrons are tossed,the total number of outcomes is $4^3 = 64$.We need the sum of the numbers on th

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$\frac{3}{32}$

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(C) Each tetrahedron has $4$ faces,so when $3$ tetrahedrons are tossed,the total number of outcomes is $4^3 = 64$.We need the sum of the numbers on the upward faces to be $5$.Let the outcomes be $(x_1, x_2, x_3)$ where $x_i \in \{1, 2, 3, 4\}$.The possible combinations that sum to $5$ are:$(1, 1, 3)$ which can occur in $3$ permutations: $(1, 1, 3), (1, 3, 1), (3, 1, 1)$.$(1, 2, 2)$ which can occur in $3$ permutations: $(1, 2, 2), (2, 1, 2), (2, 2, 1)$.Total favorable outcomes $= 3 + 3 = 6$.Therefore,the required probability $= \frac{6}{64} = \frac{3}{32}$.

The corners of regular tetrahedrons are numbered $1, 2, 3, 4$. Three tetrahedrons are tossed. The probability that the sum of the upward corners will be $5$ is

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